A question on the ultrafilter numberSurjective Maps onto $aleph$-numbersMaximal chains in a quasi-order of linear order typesIs there a modification of Martin's Axiom which admits non-measurable sets of size less than continuum?What sort of large cardinal can continuum be?Is there a locally compact, $omega_1$-compact, not $sigma$-countably compact space of size $aleph_1$?Is there nontrivial structure to forcing axioms?The minimum cardinality of an almost disjoint reaping familyThe “strong” measure numberTranslates of measure zero setThe Parovichenko cardinal, is it equal to $maxaleph_2,mathfrak p$?
Multi tool use
A question on the ultrafilter number
Surjective Maps onto $aleph$-numbersMaximal chains in a quasi-order of linear order typesIs there a modification of Martin's Axiom which admits non-measurable sets of size less than continuum?What sort of large cardinal can continuum be?Is there a locally compact, $omega_1$-compact, not $sigma$-countably compact space of size $aleph_1$?Is there nontrivial structure to forcing axioms?The minimum cardinality of an almost disjoint reaping familyThe “strong” measure numberTranslates of measure zero setThe Parovichenko cardinal, is it equal to $maxaleph_2,mathfrak p$?
$begingroup$
Let $mathfraku$ denote the ultrafilter number, which is defined to be the minimum cardinality of a subset of $mathcalP(mathbb N)$ which is a base for a nonprincipal ultrafilter on $mathbbN$. Clearly $aleph_1leq frakuleq 2^aleph_0$, so it is only interesting to study $fraku$ under the negation of CH. Kunen proved that it is consistent that CH fails and that $fraku=aleph_1$. Martin's axiom implies that $fraku=2^aleph_0$.
Is it consistent that $aleph_1<fraku<2^aleph_0$? If so, can I please have a reference?
set-theory lo.logic
asked 9 hours ago
IsaacIsaac
534
$endgroup$
add a comment |
$begingroup$
Let $mathfraku$ denote the ultrafilter number, which is defined to be the minimum cardinality of a subset of $mathcalP(mathbb N)$ which is a base for a nonprincipal ultrafilter on $mathbbN$. Clearly $aleph_1leq frakuleq 2^aleph_0$, so it is only interesting to study $fraku$ under the negation of CH. Kunen proved that it is consistent that CH fails and that $fraku=aleph_1$. Martin's axiom implies that $fraku=2^aleph_0$.
Is it consistent that $aleph_1<fraku<2^aleph_0$? If so, can I please have a reference?
set-theory lo.logic
asked 9 hours ago
IsaacIsaac
534
$endgroup$
add a comment |
$begingroup$
Let $mathfraku$ denote the ultrafilter number, which is defined to be the minimum cardinality of a subset of $mathcalP(mathbb N)$ which is a base for a nonprincipal ultrafilter on $mathbbN$. Clearly $aleph_1leq frakuleq 2^aleph_0$, so it is only interesting to study $fraku$ under the negation of CH. Kunen proved that it is consistent that CH fails and that $fraku=aleph_1$. Martin's axiom implies that $fraku=2^aleph_0$.
Is it consistent that $aleph_1<fraku<2^aleph_0$? If so, can I please have a reference?
set-theory lo.logic
asked 9 hours ago
IsaacIsaac
534
$endgroup$
Let $mathfraku$ denote the ultrafilter number, which is defined to be the minimum cardinality of a subset of $mathcalP(mathbb N)$ which is a base for a nonprincipal ultrafilter on $mathbbN$. Clearly $aleph_1leq frakuleq 2^aleph_0$, so it is only interesting to study $fraku$ under the negation of CH. Kunen proved that it is consistent that CH fails and that $fraku=aleph_1$. Martin's axiom implies that $fraku=2^aleph_0$.
Is it consistent that $aleph_1<fraku<2^aleph_0$? If so, can I please have a reference?
set-theory lo.logic
set-theory lo.logic
asked 9 hours ago
IsaacIsaac
534
asked 9 hours ago
IsaacIsaac
534
asked 9 hours ago
IsaacIsaac
534
asked 9 hours ago
IsaacIsaac
534
asked 9 hours ago
IsaacIsaac
534
534
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The answer to your question is yes. In fact, one can force to make $mathfrak u$ equal to any prescribed uncountable regular cardinal while making the cardinal of the continuum equal to any larger prescribed uncountable regular cardinal. This is proved in and old paper by Shelah and me:
Blass, Andreas(1-PAS); Shelah, Saharon(1-RTG)
Ultrafilters with small generating sets.
Israel J. Math. 65 (1989), no. 3, 259–271.
answered 7 hours ago
Andreas BlassAndreas Blass
58k7138225
$endgroup$
$begingroup$
Thanks Andreas.
$endgroup$
– Isaac
6 hours ago
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "504"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f325297%2fa-question-on-the-ultrafilter-number%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The answer to your question is yes. In fact, one can force to make $mathfrak u$ equal to any prescribed uncountable regular cardinal while making the cardinal of the continuum equal to any larger prescribed uncountable regular cardinal. This is proved in and old paper by Shelah and me:
Blass, Andreas(1-PAS); Shelah, Saharon(1-RTG)
Ultrafilters with small generating sets.
Israel J. Math. 65 (1989), no. 3, 259–271.
answered 7 hours ago
Andreas BlassAndreas Blass
58k7138225
$endgroup$
$begingroup$
Thanks Andreas.
$endgroup$
– Isaac
6 hours ago
add a comment |
$begingroup$
The answer to your question is yes. In fact, one can force to make $mathfrak u$ equal to any prescribed uncountable regular cardinal while making the cardinal of the continuum equal to any larger prescribed uncountable regular cardinal. This is proved in and old paper by Shelah and me:
Blass, Andreas(1-PAS); Shelah, Saharon(1-RTG)
Ultrafilters with small generating sets.
Israel J. Math. 65 (1989), no. 3, 259–271.
answered 7 hours ago
Andreas BlassAndreas Blass
58k7138225
$endgroup$
$begingroup$
Thanks Andreas.
$endgroup$
– Isaac
6 hours ago
add a comment |
$begingroup$
The answer to your question is yes. In fact, one can force to make $mathfrak u$ equal to any prescribed uncountable regular cardinal while making the cardinal of the continuum equal to any larger prescribed uncountable regular cardinal. This is proved in and old paper by Shelah and me:
Blass, Andreas(1-PAS); Shelah, Saharon(1-RTG)
Ultrafilters with small generating sets.
Israel J. Math. 65 (1989), no. 3, 259–271.
answered 7 hours ago
Andreas BlassAndreas Blass
58k7138225
$endgroup$
The answer to your question is yes. In fact, one can force to make $mathfrak u$ equal to any prescribed uncountable regular cardinal while making the cardinal of the continuum equal to any larger prescribed uncountable regular cardinal. This is proved in and old paper by Shelah and me:
Blass, Andreas(1-PAS); Shelah, Saharon(1-RTG)
Ultrafilters with small generating sets.
Israel J. Math. 65 (1989), no. 3, 259–271.
answered 7 hours ago
Andreas BlassAndreas Blass
58k7138225
answered 7 hours ago
Andreas BlassAndreas Blass
58k7138225
answered 7 hours ago
Andreas BlassAndreas Blass
58k7138225
answered 7 hours ago
Andreas BlassAndreas Blass
58k7138225
58k7138225
$begingroup$
Thanks Andreas.
$endgroup$
– Isaac
6 hours ago
add a comment |
$begingroup$
Thanks Andreas.
$endgroup$
– Isaac
6 hours ago
$begingroup$
Thanks Andreas.
$endgroup$
– Isaac
6 hours ago
$begingroup$
Thanks Andreas.
$endgroup$
– Isaac
6 hours ago
add a comment |
Thanks for contributing an answer to MathOverflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f325297%2fa-question-on-the-ultrafilter-number%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
BfRul,TdVBpGnnegdqj,l3UZwSdZD98zjAZBzU6en1YQ6g6bKOeTWFJTEx22QH3ze6Q,L9 IbDLYtVFOgCfUQsgM2PhGiQSDK,tKm3h
